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Question Number 205173 by mr W last updated on 12/Mar/24

$$find$$$$S=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(a^2+n^2)}^2}$$

Commented by lepuissantcedricjunior last updated on 12/Mar/24

$$\mathit{S}=\underset{\mathit{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{({\mathit{a}}^2+{\mathit{n}}^2)}^2}={\int }_0^{\mathrm{\infty }}\frac{\mathit{d}\mathit{x}}{{({\mathit{a}}^2+{\mathit{x}}^2)}^2}$$$$\mathit{S}={\int }_0^{+\mathrm{\infty }}\frac{\mathit{d}\mathit{x}}{{\mathit{a}}^4{[1+{\left(\frac{\mathit{x}}{\mathit{a}}\right)}^2]}^2}$$$$\mathit{p}\mathit{o}\mathit{s}\mathit{o}\mathit{n}\mathit{s}\frac{\mathit{x}}{\mathit{a}}=\mathit{t}\mathit{a}\mathit{n}\mathit{\beta }\iff \mathit{d}\mathit{x}=\mathit{a}(1+\mathit{t}\mathit{a}\stackrel{2}{\mathit{n}\beta })\mathit{d}\mathit{\beta }$$$$\mathit{q}\mathit{d}:\{\begin{array}{l}\mathit{x}\to 0\\ \mathit{x}\to \mathrm{\infty }\end{array}\Rightarrow \{\begin{array}{l}\mathit{\beta }\to 0\\ \mathit{\beta }\to \frac{\mathit{\pi }}{2}\end{array}$$$$\mathit{S}={\int }_0^{\frac{\mathit{\pi }}{2}}\frac{\mathit{d}\mathit{\beta }}{{\mathit{a}}^3(1+\mathit{t}\mathit{a}\stackrel{2}{\mathit{n}\beta })}={\int }_0^{\frac{\mathit{\pi }}{2}}\frac{\mathit{c}\mathit{o}\stackrel{2}{\mathit{s}\beta }}{{\mathit{a}}^3}\mathit{d}\mathit{\beta }$$$$\mathbf{S}=\frac{1}{{\mathit{a}}^3}{[\frac{1}{2}\mathit{\beta }+\frac{1}{4}\mathit{s}\mathit{i}\mathit{n}2\mathit{\beta }]}_0^{\frac{\mathit{\pi }}{2}}$$$$\mathbf{S}=\frac{\mathit{\pi }}{4{\mathit{a}}^3}$$$$$$

Commented by mr W last updated on 12/Mar/24

$$howcanyouget$$$$\underset{\mathit{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{({\mathit{a}}^2+{\mathit{n}}^2)}^2}={\int }_0^{\mathrm{\infty }}\frac{\mathit{d}\mathit{x}}{{({\mathit{a}}^2+{\mathit{x}}^2)}^2}?$$

Answered by Berbere last updated on 13/Mar/24

$$\sum _{n⩾1}\frac{1}{(a^2+n^2)}=f\left(a\right)$$$$f\left(a\right)=\frac{1}{2ai}(\sum _{n⩾1}\frac{1}{n-ia}-\frac{1}{n+ia})=\frac{1}{2ai}(\sum _{n⩾0}{\int }_0^1{t}^{n-ia}-t^{n+ia})$$$$=\frac{1}{2ai}{\int }_0^1\frac{t^{-ia}-t^{ia}}{1-t}dt;\mathrm{\Psi }\left(z\right)=-\gamma +{\int }_0^1\frac{1-x^{z-1}}{1-x}dx$$$$S=\frac{1}{2ai}(\mathrm{\Psi }(1+ia)-\mathrm{\Psi }(1-ia))$$$$=\frac{1}{2ai}(\frac{1}{ia}+\mathrm{\Psi }\left(ia\right)-\mathrm{\Psi }(1-ia))$$$$\mathrm{\Psi }(1-z)-\mathrm{\Psi }\left(z\right)=\pi cot\left(\pi z\right)$$$$=-\frac{1}{2a^2}-\frac{1}{2ai}\pi cot\left(i\pi a\right)=-\frac{1}{2a^2}+\frac{\pi }{2a}coth\left(\pi a\right)$$$$\sum _{n⩾1}\frac{1}{(n^2+a^2)}⩽\zeta \left(2\right)serieCvunifomly\Rightarrow$$$$a\stackrel{f}{\to }\sum _{n⩾1}\frac{1}{(n^2+a^2)}isderivable?wecanchange\mathrm{\Sigma }andderivation$$$$f^{\prime }\left(a\right)=\sum _{n⩾1}\frac{-2a}{{(n^2+a^2)}^2}=\frac{1}{a^3}-\frac{\pi }{2a^2}coth\left(\pi a\right)+\frac{{\pi }^2}{2a^2}.\frac{1}{{sinh}^2\left(a\right)}$$$$\sum _{n⩾1}\frac{1}{{(n^2+a^2)}^2}=-\frac{1}{2a^4}+\frac{\pi }{4a^3}coth\left(\pi a\right)+\frac{{\pi }^2}{4a^2{sinh}^2\left(\pi a\right)}$$$$$$

Commented by mr W last updated on 13/Mar/24

$$thankssir!$$$$nicesolution!$$

Commented by Berbere last updated on 13/Mar/24

$$YessirWithepleasur$$$$ilostmyAccuntWitcher$$$$Godblessyou$$

Commented by mr W last updated on 13/Mar/24

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